What is the Work and Friction in This Incline Scenario?

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A worker pushes a 93N crate up a 37-degree incline with a force of 85N, resulting in 850 joules of work done by the worker. The work done by gravity is calculated as 560 N, but the calculations for work done by friction and the coefficient of friction remain unclear. The net work done by non-conservative forces, including the worker's force and friction, is suggested to be equal to the negative change in energy of the box. Clarification is sought on whether the force of friction is -850 joules and if the normal force is indeed 93 N. Understanding these concepts is crucial for solving the remaining parts of the problem.
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Homework Statement



A worker pushes a 93N crate 10m up a 37 degree incline with constant velocity. The worker exerts a force of 85N parallel to the incline.

a. How much work does he do?
b. How much work is done by gravity?
c. How much work is done by friction?
d. What is the coefficient of friction between the box and the incline?

Homework Equations



Work = Force x Distance
Ff = (u)(Fn)

The Attempt at a Solution



Part A:

work = (85 N)(10 m) = 850 joules

Part B:

work = (10 * sin(37)) * 93 N = 560 N

I believe these are correct. The parts that I do not know how to do are C and D. Anyone know how? I greatly appreciate all help :)
 
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HINT:

The net work done by non conservative forces (i.e. the force of the push and friction) will equal the negative of the change in energy of the box:

W_{non-conservative}=-\Delta E

Can you use this to find the answers for parts C and D?
 
Thank you for your help!

So the force of friction is -850 joules?

For Part D, I am supposed to use the friction force for Ff and the normal force for Fn, right?

Is the normal force 93 N?
 
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